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13.5

A Closer Look

Colligative Properties of Electrolyte

Solutions

(van’t Hoff Factor i)

• Water is known as “the universal solvent”

• This is due to water’s unusual properties, it is:

1. Bent in shape

2. A Highly polar molecule

3. And it causes ionic and polar substances to be soluble,

(remember “like dissolves like”)

4. Most ions are soluble in water: cations separate from anions

• A solution is homogeneous mixture of solute

and solvent

– Solutions may be gases, liquids, or solids

• Solvent = component in largest amount

• Solutes = other components

• Intermolecular forces become arranged in the

process of making solutions with condensed

phases.

• Water molecules orient

themselves on the NaCl

crystals

• H-bonds between waters are

broken

• NaCl dissociates into Na+ and

Cl-

• Ion-dipole forces from

between the Na+ and the

negative end of the water

dipole

– Similar between Cl- and positive

end of water

• Solvation: interaction

between solvent and solute

– Hydration: If water is the solvent

• Covalent compounds (like alcohols) do not

separate into ions, they stay together

because they are covalently bonded.

• Alcohols become less soluble because of the

parent chain, as more carbons are added it

becomes less polar (less like water).

• This causes less hydrogen bonding between

the alcohol and the water.

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Electrolytes are ionic, nonelectrolytes are covalent• The van’t Hoff factor = i

• Ionic compounds have a

factor greater than one:

• NaCl = 2 = 1 Na+ and

1 Cl-

• H2SO4 = 3 = 2 H+ and

1 SO42-

• CH3OH = 1

(covalent, does NOT dissociate)

Memorize!

Solution Formation, Spontaneity,

Disorder, & Chemical

Reactions

• A spontaneous process occurs without outside intervention ∆∆∆∆S

• When the energy of a system decreases, the process is spontaneous

• Spontaneous tends to be exothermic

– Some move from lower to higher energy state (endothermic reaction)

• Entropy is the amount of

disorder in the system

– In most cases, solution

formation is favored by

the increase in entropy

that accompanies mixing

13.3

Factors Affecting Solubility• Pairs of liquids that mix in any proportions are

said to be miscible.

• Liquids that do not mix significantly are immiscible.

• Intermolecular forces (IMF) are important in determining solubility.• The stronger the attraction between solute and solvent the

greater the solubility

• Network solids do not dissolve due to strong intramolecular forces within the solid.

• Note “Rhonda’s Handout” Next slide…

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13.4

Ways of Expressing Concentration

• Solution: is a

homogeneous mixture, the

components are uniformly

intermingled.

• Solvent: the substance

present in the largest

amount (usually the

component doing the

dissolving).

• Solute: the other

substance(s) (usually the

component being

dissolved).

• Aqueous solutions: are

solutions with water as the

solvent.

Common Metric Conversions

WATCH UNITS!

1000 mL = 1 L

Molarity (M)

• The concentration of a solute is a measure of the amount of solute in a given volume of solution.

• Molarity will change with a change in temperature (Volume)

• Concentration can have various units, Molarity (M) is commonly used by chemists:

Molarity (M) = moles of solute = mol

Liters of solution L

• Calculate the standard solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

• 11.5 g NaOH x 1 mol NaOH = 0.288 mol

40.0 g NaOH

• M = mol = 0.288 mol = 0.192 M NaOH

L 1.50 L

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Concentration of Ions

• If a solution is 1.0 M NaCl it contains 1.0 mol

Na+ ions and 1.0 mole of Cl- for example…

• Give the concentration of each type of ion in

the following solution:

a. 0.50 M Co(NO3)2

Answer: Co(NO3)2 � Co2+ + 2 NO3-

1 mol Co2+ ions = 0.50 M Co2+

2 mol NO3- ions = 2 (0.50 M) NO3- = 1.0 M NO3-

Calculate the number of moles of Cl- ions in 1.75

L of 1.0 x 10-3 M ZnCl2

• Remember: M * L = mol

• ZnCl2 � 1 mol of Zn2+ + 2 mols of Cl-

Calculate the number of moles of Cl- ions in 1.75

L of 1.0 x 10-3 M ZnCl2

• M * L = mol

• ZnCl2 � 1 mol of Zn2+ + 2 mols of Cl-

• (1.0 x 10-3 M)(1.75 L) = 0.00175 moles

• (0.00175 moles)(2 mol Cl-) = 0.00350 mol Cl-

• The dilution process is used to save time and space in the lab, often concentrated solutions (called stock solutions) are diluted by adding water to achieve the desired molarity.

M1 * V1 = M2 * V2

• In performing calculations associated with

dilutions remember that the moles of a solute are

not changed by the dilution:

M1 * V1 = M2 * V2

M1 * V1 = mol solute before dilution = mol solute before dilution M2 * V2

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• M1 * V1 = M2 * V2

• How many mL of 6.00 M NaOH solution are necessary to prepare 300. mL of 1.20 M NaOHsolution?

• M1 * V1 = M2 * V2

• How many mL of 6.00 M NaOH solution are necessary to prepare 300. mL of 1.20 M NaOHsolution?

• (6.00 M)(x mL) = (1.20 M)(300. mL)

• 6.00 (x) = 360.

6.00 6.00

x = 60.0 mL

Mole Fraction

• Based on the number of moles of one or more components.

• Remember mass converts to moles using molar mass (amu)

• Mole fraction has no units

• Fractions range from 0 to 1

Summary of

Colligative Properties

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